Suppose there is a fair six faced dice, and you roll it once at each round. The rule is as follows:
1). If you roll 1 or 2 at an odd-number round, you win the game and the game ends immediately.
2). If you roll 1 or 2 at an even-number round, you lose the game and it ends immediately.
If you roll 3 at any round, then it is a draw and the game ends immediately. If two consecutive $3$'s appear then the game ends immediately in a draw.
4). Otherwise the game goes on.
Now, what is the probability that you will win the game?
I attempted to compute the probability of winning at the 1,3,5,...-th round separately and sum them up. However, due to rule 3) it is ever more complicated to compute this for large round numbers.
Another attempt was to calculate the probability of stopping at each round and to get a closed form expression for this sequence, but this is also hard, because it depends backwards on the previous rounds in an endless manner.
Simply put, if we only consider 1) and 2), or only consider 3), the problem isn't hard, but a combination of the three really messes things up.
Can anyone help? Thanks in advance.
EDIT: I'm very sorry for the mistake when I transcribed rule 3), I have corrected it.